Given `E^(@)(Zn^(2+)//Zn)=-0.76V`
`E^(@)(Ni^(2+)//Ni)=-0.25V`
Calculate the EMF of the cell where the following reaction is taking place
`Zn_((s))+Ni_((aq))^(2+)toZn_((aq))^(2+)+Ni_((s))`

Options

A. `0.51V`
B. 1.01V
C. `-0.51V`
D. 0.25V

Explanation

To calculate the EMF of the cell for the reaction \( \text{Zn}_{(s)} + \text{Ni}^{2+}_{(aq)} \rightarrow \text{Zn}^{2+}_{(aq)} + \text{Ni}_{(s)} \), we can follow these steps:

### Step 1: Identify the half-reactions
The overall cell reaction can be broken down into two half-reactions:
1. Oxidation half-reaction (at the anode):
\[ \text{Zn}_{(s)} \rightarrow \text{Zn}^{2+}_{(aq)} + 2e^- \]
2. Reduction half-reaction (at the cathode):
\[ \text{Ni}^{2+}_{(aq)} + 2e^- \rightarrow \text{Ni}_{(s)} \]

### Step 2: Write the standard reduction potentials
From the question, we have the standard reduction potentials:
- For \( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}_{(s)} \):
\[ E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76 \, \text{V} \]
- For \( \text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}_{(s)} \):
\[ E^\circ(\text{Ni}^{2+}/\text{Ni}) = -0.25 \, \text{V} \]

### Step 3: Determine the anode and cathode
- The anode is where oxidation occurs, which is the Zn half-reaction.
- The cathode is where reduction occurs, which is the Ni half-reaction.

### Step 4: Calculate the EMF of the cell
The EMF of the cell can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = E^\circ(\text{Ni}^{2+}/\text{Ni}) - E^\circ(\text{Zn}^{2+}/\text{Zn}) \]
\[ E^\circ_{\text{cell}} = (-0.25 \, \text{V}) - (-0.76 \, \text{V}) \]
\[ E^\circ_{\text{cell}} = -0.25 \, \text{V} + 0.76 \, \text{V} \]
\[ E^\circ_{\text{cell}} = 0.51 \, \text{V} \]

### Final Answer
The EMF of the cell is \( 0.51 \, \text{V} \).

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